IB HL Math Homework #3 Solutions
Assigned: 9/10/07, Monday
Due: 9/13/07, Thursday
1) Find the value of (1  3i ) 7 using DeMoivre's Theorem.
(1  3i ) 7  (2cis (
4 7
28
 4
))  2 7 cis (
)  128(cis 
3
3
 3

)  64  64 3i

2) Find the four solutions to the following equation: z 4  128  128 3i .
z 4  128  128 3i
2
z 4  256cis (
 2k )
3
2
z  (256cis (
 2k ))1 / 4
3
2 2k
z  2561 / 4 cis (

), k  0,1,2,3
12
4
 k
z  4cis (  ), k  0,1,2,3
6 2

2
7
5
z  4cis ( ) , or , z  4cis ( ) , or z  4cis ( ) , or z  4cis ( )
6
3
6
3
z  2 3  2i , or z  2  2 3i , or z  2 3  2i , or z  2  2 3i
3) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic
equation with a leading coefficient of 1 with the roots 2r and 2s.
We know that r + s = 5 and rs = 3. It follows that 2r + 2s = 2(5) = 10 and (2r)(2s) =
4(3) = 12. Thus, the desired quadratic is x2 – 10x + 12 = 0.
4) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic
equation with a leading coefficient of 1 with the roots r2 and s2.
r2+s2 = (r+s)2 – 2rs = 52 – 2(3) = 19.
r2s2 = (rs)2 = 32 = 9
Thus, the desired quadratic equation is x2 – 19x + 9 = 0.
5) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic
equation with a leading coefficient of 1 with the roots r3 and s3.
r3+s3 = (r+s)3 – 3r2s – 3rs2 = (r+s)3 – 3rs(r+s) = 53 – 3(3)5 = 125 – 45 = 80.
r3s3 = (rs)3 = 33 = 27,
so the desired quadratic equation is x2 – 80x + 27 = 0.
6) Let the roots of the quadratic equation x2 – 5x + 3 = 0 be r and s. Find the quadratic
equation with a leading coefficient of 1 with the roots 1/r and 1/s.
1 1 sr 5
 

r s
rs
3
7) The cubic equation x3 – 7x2 + 17x – 15 = 0 has the root 2 + i. What are the other two
roots?
Another root must be 2 – i. These sum to 4. The sum of the roots is 7, thus the last
root is 3. The three roots are 2 + i, 2 – i, and 3.